Having landed on a newly discovered planet, an astronaut sets up a simple pedulum of length 1.29 m and finds that it makes 567 oscillations in 1020 s. The amplitude of the oscillations is very small compared to the pendulum's length.
What is the gravitational acceleration on the surface of this planet? (Answer in units of m/s squared)
Use the equation:
Period = 2 pi sqrt (L/g') = 1020/567 s
= 1.799 s
Solve for the acceleration of gravity on that planet, g'. It will be different from the g that exists on Earth.
L is the pendulum length
sqrt(L/g') = 1.799/(2 pi) = 0.2863 s
L/g' = 0.08197 s^2
g'/L = 12.2 s^-2
g' = 15.7 m/s^2
To find the gravitational acceleration on the surface of the planet, we can use the formula for the period of a simple pendulum:
T = 2π√(L/g)
Where:
T is the period of the pendulum
L is the length of the pendulum
g is the gravitational acceleration
From the given information, we know:
Length of the pendulum, L = 1.29 m
Number of oscillations, N = 567
Total time for the oscillations, t = 1020 s
First, let's calculate the period of the pendulum, T:
T = t / N
T = 1020 s / 567
Now, we can substitute the values of T and L into the formula mentioned earlier and solve for g:
T = 2π√(L/g)
1020 s / 567 = 2π√(1.29 m / g)
To solve for g, we need to isolate it on one side of the equation. Let's square both sides of the equation to eliminate the square root term:
(1020 s / 567)² = (2π√(1.29 m / g))²
Simplifying:
(1020 s / 567)² = (2π)² * (1.29 m / g)
Now, isolate g:
g = (2π)² * (1.29 m) / ((1020 s / 567)²)
Using a calculator, compute the value:
g ≈ 8.05 m/s²
Therefore, the gravitational acceleration on the surface of this newly discovered planet is approximately 8.05 m/s².